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The imaginary part of $i^{i}$ is
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Verified Answer
The correct answer is:
0
We have,
$$
e^{i x}=\cos x+t \sin x
$$
Put $x=\frac{\pi}{2}$
$$
\begin{aligned}
&\Rightarrow \quad e^{i \frac{\pi}{2}}=\cos \frac{\pi}{2}+i \sin \frac{\pi}{2} \\
&\Rightarrow \quad e^{i \frac{\pi}{2}}=i \Rightarrow e^{i^{2} \frac{\pi}{2}}=\langle i)^{i} \Rightarrow i=e^{-\frac{\pi}{2}}
\end{aligned}
$$
$\therefore$ Imaginary part of $i$ is zero.
$$
e^{i x}=\cos x+t \sin x
$$
Put $x=\frac{\pi}{2}$
$$
\begin{aligned}
&\Rightarrow \quad e^{i \frac{\pi}{2}}=\cos \frac{\pi}{2}+i \sin \frac{\pi}{2} \\
&\Rightarrow \quad e^{i \frac{\pi}{2}}=i \Rightarrow e^{i^{2} \frac{\pi}{2}}=\langle i)^{i} \Rightarrow i=e^{-\frac{\pi}{2}}
\end{aligned}
$$
$\therefore$ Imaginary part of $i$ is zero.
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